anonymous
  • anonymous
find the domain of the function f(x)= 1/sqrt (4x-8) please show calculations
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
huh
anonymous
  • anonymous
callum can you show the steps please
anonymous
  • anonymous
(2, infinity)

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anonymous
  • anonymous
\[f(x) = \frac{1}{\sqrt{4x-8}}\] now you want to look at conditions on 4x-8. What can x reasonably be? Well you cannot take a square root of a negative number, and the denominator cannot be zero, so we need: \[4x-8 > 0\] \[x>\frac{1}{2}\] So I was wrong above, the domain is actually: \[\left(\frac{1}{2}, \infty \right)\]
anonymous
  • anonymous
x > 2
anonymous
  • anonymous
think you divided the wrong side there
anonymous
  • anonymous
oh yes sorry it's 2 lol, quite right. \[(2,\infty)\]
anonymous
  • anonymous
rickjbr can you demonstrate how you got your answer please
anonymous
  • anonymous
you'll have imaginary numbers if the square root is allowed to go negative
anonymous
  • anonymous
thus not part of the real number system
anonymous
  • anonymous
callum was right in his calculation 4x - 8 > 0 4(x-2) > 0 x-2 > 0 x > 2
anonymous
  • anonymous
yeah my reasoning was right, it's just I divided the inequality wrong and got x>1/2. rickjbr corrected this to x>2
anonymous
  • anonymous
thanks for the help to both of your
anonymous
  • anonymous
np
anonymous
  • anonymous
how about this. 2x^2-3x+5...find and simplify the difference quotient
anonymous
  • anonymous
The question reads Find and simplify the difference quotient f(x+h)-f(x)/h for the given function f(x)=2x^(2)-3x+5
anonymous
  • anonymous
ah, right
anonymous
  • anonymous
(2(x+h)^2-3(x+h)+5-2x^2+3x-5)/h
anonymous
  • anonymous
(2(x^2+2xh+h^2)-3(x+h)+5-2x^2+3x-5)/h
anonymous
  • anonymous
(2x^2+4xh+2h^2-3x-3h+5-2x^2+3x-5)/h
anonymous
  • anonymous
(2h^2+4xh+3x)/h
anonymous
  • anonymous
is that the final answer
anonymous
  • anonymous
checking, a bit rusty
anonymous
  • anonymous
my options are a)4x+2h, b)4x+2h-3, c)4h+2x-3, d)4h+2x, e)4x+4h
anonymous
  • anonymous
i'll redo it, might've missed something
anonymous
  • anonymous
ok
anonymous
  • anonymous
ah, silly
anonymous
  • anonymous
that last step was it. instead of (2h^2+4xh+3x)/h it should be (2h^2+4xh-3h)/h
anonymous
  • anonymous
simply factor out an h
anonymous
  • anonymous
[h(2h+4x-3)]/h =2h+4x-3
anonymous
  • anonymous
that's it
anonymous
  • anonymous
thanks greatly. i dont know if you have the time...but im reviewing for a test and I have a whole lot of other questions. if you can't help me, i understand, i will just post on the main forum
anonymous
  • anonymous
probably better off posting on main forum, if i can help, i'll try
anonymous
  • anonymous
ok thanks for the help
anonymous
  • anonymous
Find \[\frac{f(x+h)-f(x)}{h}\] for \[f(x)=2x^2-3x+5\] \[\frac{f(x+h)-f(x)}{h} = \frac{(2(x+h)^2-3(x+h)+5) - (2x^2-3x+5)}{h}\] \[= \frac{(2x^2+4xh+2h^2-3x-3h)-(2x^2-3x)}{h}\] \[=\frac{4xh+2h^2-3h}{h} = 4x+2h-3\] We can verify that this is right by noting that \[f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h) - f(x)}{h}\] The left hand side of this is equal to the derivative of f, which we can just do right here: \[f'(x) = 4x - 3\] and we should get that this equals the limit as h goes to zero of \[\frac{f(x+h) - f(x)}{h}.\] We have already calculated that \[\frac{f(x+h) - f(x)}{h} = 4x+2h-3\] so taking the limit of this as h goes to zero we get: \[\lim_{h\rightarrow 0}(4x+2h - 3) = 4x+0-3 = 4x-3.\] Hence we ave verified that the simplification has been done correctly.

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